Explore topic-wise InterviewSolutions in Current Affairs.

This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.

1.

In an aquose solution of D-Glucose the percentage of alpha and beta anomers at equilibrium condition are respectively

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80 and 20
20 and 80
36 and 64
64 and

ANSWER :C
2.

In an aqueous solution the ionisation constants for carbonic acid are K_(1)=4.2xx10^(-7)andK_(2) =4.8xx10^(-11). Select the correct statement for a saturated 0.034 M solution of the carbonic acid.

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The concentration of `CO_(3)^(2-)` is 0.034 M
The concentration of `CO_(3)^(2-)` is GREATER than that of `HCO_(3)^(-)`
The concentration of `H^(+)andHCO_(3)^(-)` are approximately equal
The concentration of `H^(+)` is double that of `CO_(3)^(2-)`

Solution :`K_(1)gtgtK_(2)`, i.e., SECOND dissociation is practically negligible
3.

In an aqueous solution of urea, mole fraction of urea is 0.4. Then w/w% of urea is:

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0.155
0.31
0.345
0.69

Solution :Mole FRACTION of urea =0.4
0.4 MOLES of urea +0.6 moles of `H_(2)O`
`%(w//w)=(0.4xx60)/((0.4xx60)+(0.6xx18))=(24)/(24+10.8)XX100`
`=(24)/(34.8)xx100=69%`
4.

In an aqueous solution, hydrogen (H_(2)) will not reduce:

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`Fe^(3+)`
`Cu^(2+)`
`ZN^(2+)`
`AG^(+)`

Solution :Only `Zn` and `Fe` are above `H`, ALSO `Fe^(3+)` can be REDUCED to `Fe^(2+)` by `H`.
5.

In an aqueous solution, how does specific conductivity of electrolytes change with addition of water?

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Solution : SPECIFIC conductivity decreases because the number of IONS PER unit VOLUME decreases.
6.

In an aqueous solution containing Cu^(2+), Ag^(+), Sn^(2+) and Hg_(2)^(2+) ions, the concentration of each ion is 1 (M). Electrolysis of the solution is carried out in presence of inert electrodes. Given: E_(Hg_(2)^(2+)|Hg)^(@)=+0.79V and E_(Sn^(2+)|Sn)^(@)=-0.14V. If the voltage applied on the electrodes is gradually increased, then the metal that will deposit first and that at the end of the process at the cathode, respectively, are -

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SN, Ag
Cu, Hg
Cu, Sn
Ag, Sn

Answer :D
7.

In an aqueous solution how does specific conductivity of electrolytes change with addition of water?

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SOLUTION :Conductivity decreases because NUMBER of IONS per unit VOLUME decreases.
8.

In an aqueous solution, how does specific conductivity of electrolytes change with addition of water ?

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Solution :* SPECIFIC CONDUCTIVITY of solution decreases with DILUTION of solution.
* On addition of water to the electrolytic solution, i.e., on dilution the interionic distance between ions get increases. So, number of ions in specific volume of solution get decreases. and hence, specific conductivity get decreases on dilution.
9.

In an aqueous of volume500 ml when the reaction, 2Ag^(+) + Cu rarr Cu^(2+) + 2Ag, reaches equilibrium, the concentration of Cu^(2+) ions is x molar: To this solution 500ml of water is added. At the new equilibrium the concentration of Cu^(2+) ions would be

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2X molar
between XAND 2x molar
less than `(1)/(2) X `molar
between 0.005x molar and 0.5x molar

Answer :C,D
10.

In an analytical determination of aresenic , a solution containing aresenious acid, (H_(3)AsO_(3)). KI and a small amount of starch is electrolysed. The electrolysis produes free I_(2) from I^(-) ion and the I_(2) immediately oxidises the arsenious acid to hydrogen arsenate ion, (HAsO_(4)^(2-)). I_(2(aq.))+H_(3)AsO_(3(aq.))+H_(2)O_((l))rarr2I_((aq.))^(-)+HAsO_(4(aq.))^(2-)+4H_((aq.))^(+) When the oxidation of arsenic is complete, the free iodine combines with thestarch to give a deep blue colour.If during a particular run, it takes 65.3 s for a current of 10.5 mA to give an end point (indicated by the blue colour), how many grams of arsenic and H_(3)AsO_(3) are present in the solution? ("At. wt. of "As = 75)

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ANSWER :As `= 2.66 XX 10^(-4)g`,
`H_(3)AsO_(3) = 5.0 xx 10^(-4)g ;`
11.

In an antiflourite structure, cations occupy

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OCTAHEDRAL VOIDS
centre of cube
tetrahedral voids
CORNERS of cube

Answer :C
12.

In an answer paper a student wrote as "carborundum crystals are very soft." What is your opinion?

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SOLUTION :CARBORUNDUM CRYSTALS are very HARD.
13.

In an answer paper a student wrote as "carborundum crystals are very soft." Do you agree with this?

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ANSWER :no
14.

In an answer paper a student wrote as "carborundum crystals are very soft." In which crystal type carborundum is included?

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SOLUTION :Covaient CRYSTAL.
15.

In an amino acid, the carboxyl group ionises at pK_(a_1)=2.34 and ammonium ion at pK_(a_2)=9.60 . The isoelectric point of the amino acid is at pH

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5.97
2.34
`9.60`
`6.97`

Solution :ISOELECTRIC POINT (pH) `=(pK_(a_1) + pK_(a_2))/2 =(2.34+9.60)/2`=5.97
16.

In an amino - acid , the carboxyl group ionises at pKa_(1) = 2.34 and the ammonium ion at pKa_(2) = 9.60 . The isoelectric point of the amino acid is at pH ………………..

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5.97
2.34
9.6
6.97

Answer :A
17.

In an adsorption expetiment, a graph between log (x/m) varuse log P was found to be linear with a slope of 45^(@). The intercept on the log (x/m) axis was found to be 0.3010. Calculate the amount of the gas adsorbed per gram of charcoal under a pressure of 0.5 atmosphere.

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Solution :According to Freundlich equation, `x/m =k P^(1//) or log ""(x)/(m)=log k+1/n log P`
`THEREFORE` Plot of `x/m` varsus log P is linear with SLOPE `=1/n` and INTERCEPT `=log k.` Thus,
`1/n = TNA theta = TAN 45^(@)=1 or n=1`
`log k =0.3010 or k= "Antilig" 0.3010=2`
At `P =0.5 atm, x/m =k P^(1//n)=2xx(0.5)^(1)=1.0`
18.

In an adsorption experiment a graph between log x/m vs. log p is found to be linear with a slope of 45^(@). The intercept on the log x/m axis was found to be 0.3010. What is x/m If presure is 0.6 bar (tan 45^(@) = 1)

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0.6
1.2
2.4
0.3

Answer :B
19.

In an alkalinemedium acetaldehyde undergoes :

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Benzoin condensation
Aldol condensation
Polymerisation
Cannizzaro.s reaction

ANSWER :B
20.

In an adsorption experiment, a graph of log(x/m) versus log P was found to be linear with a slope of 45°, and the the intercept of of 0.3010. The amount of gas adsorbed per gram charcoal under a pressure of 0.8 bar is

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1.2
1.4
1.6
1.8

Answer :C
21.

In an Adsorption experiment a graph between log x/m versus log P was found to be linear with a slope of 45^(@) the intercept of the log x/m was found to be 0.3010. Calculate the amount of gas adsorbed per gram of charcoal under a pressure of 0.6 bar.

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SOLUTION :`1.2`
22.

In an adsorption experiment a graph between log x/m vs. log p is found to be linear with a slope of 45^@ . The intercept on the log x/m axis was found to be 0.3010. What is x/m if pressure is 0.6 bar

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0.6
1.2
2.4
0.3

Answer :B
23.

In an adiabatic process:

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The system EXCHANGES HEAT with surrounding
Pressure is maintained constant
There is perfect heat insulation
The GAS is isothermally expanded

Answer :C
24.

In an adiabatic expansion of an ideal gas.

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`DELTA T = 0`
W = 0
q = 0
`Delta U = 0`.

Answer :C
25.

In an acidic medium ………………….. behaves as the strongest base (nitrobenzen. Aniline, phenol),

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ANSWER :ANILINE
26.

In an acidified solution of K_2Cr_2O_7, H_2O_2 is added. Then

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solution TURNS green due to formation of `Cr_2O_3`
solution turns yellow due to formation of `K_2CrO_4`
a blue coloured compound `CrO(O_2)_(2)` is FORMED
solution gives green precipitate of `Cr(OH)_(3)`

Answer :C
27.

In Amylpectin the linkage absent is

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`C_1 & C_4`
`C_1 & C_6`
`C_1 & C_2`
Both `C_1 & C_6` and `C_1 &C_4`

ANSWER :C
28.

In amylopectin glycosidic long chain and branching occurs in between

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C-1 of ONE `ALPHA`-D-glucopyranos`to`C-4 of ANOTHER `alpha`-D-GLUCOPYRANOSE and BRANCHING at C-1 of one glucopyranose C-6 of another glucopyranose

Answer :A
29.

In amylopectin glycosidic branching present in between

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1-4 `ALPHA`-D-glucopyranose
1-4-`BETA`-D-glucopyranose
1-6-`alpha`-D-glucopyranose
1-6 `beta`-D-glucopyranose

Answer :C
30.

In amixure of NaOH and NaCl 50% sodium is present. Calculate mass % of NaOH in the solution.

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`95.2%`
0.5875
0.1567
0.227

Solution :In 100gm MIXTURE contain `W_(Na)=50GM`
`n_(NAOH)=a, n_(NaCl)=b`
40a+58.5b=100 . . . .(i)
`axx23+bxx23=50gm`
`a+b=(50)/(23)` . . . .(ii) `40a+40b=(50xx40)/(23)`
40a+58.5b=100
`18.5b=100-(200)/(23)`
`b=(100-86.9)/(18.5)=0.7`.
`a=(50)/(23)-0.7=2.17-0.7`
In 100 gm mixture `rArr W_(NaOH)=58.8rArr58.8%` APPROX.
31.

In aluminothermic process aluminium acts as :

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OXIDISING AGENT
REDUCING agent
flux
None of these

ANSWER :B
32.

In aluminium extraction by the Baeyer's process, alumina is extracted from bauxite by sodium hydroxide at high temperatures and pressures. Al_(2)O_(3(s))+2OH_((aq))^(-) to 2AlO_(2(aq))^(-)+H_(2)O_(l) Solid impurities such as Fe_(2)O_(3) and SiO_(2) are removed and then Al(OH)_(4)^(-) is reprecipitated. In the industrial world

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carbon dioxide is ADDED to precipitate the alumina
temperature and pressure are dropped and the supersaturated SOLUTION seeded
both (a) and (B) are practised
the WATER is evaporated

Solution :Both (a) and (b) are practised.
33.

In aluminothermic process, Al is used as....

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reducing AGENT 
OXIDISING agent 
CATALYST 
electrolyte 

ANSWER :A
34.

In alumino-thermy, aluminium is heated with :

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CALCIUM oxide
Chromium oxide
Magnesium oxide
Sodium oxide

Answer :B
35.

In aluminium oxide, the oxide ions are arranged in hexagonal close packed (hcp) arrangement and the aluminium occupy 2//3 of octahedral voids. What is the formula of oxide ?

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Solution :Number of octahedral sites corresponding to each atom in hcp=1
Number of OXYGEN (O) atoms = 1
It is GIVEN that `2//3` of the octahedral sites are OCCUPIED by aluminium ions.
Hence, for each oxide ion, there will be `2//3` aluminium ions. so, number of AL atoms = `2//3`
Hence, the formula of the oxide = `Al_(2//3) O_1` or `Al_2O_3`
36.

In Alum: K_(2)SO_(4), Al_(2)(SO_(4))_(3). 24 H_(2)O Which metal can replace Al

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Cr
Mn
In
SC

Solution :`K_(2)SO_(4), Cr_(2)(SO_(4))_(3)24H_(2)O` CHROME alum can FORM.
37.

In aluminates, the coordination number of Al is

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`O_(2)`
`CO_(2)`
HELIUM
All of these

Solution :In ARTIFICIAL repiration a mixture of `O_(2)` with `CO_(2)` or helium is used. Helium is used as it is less soluble in blood.
38.

In aluminates coordination number of Al is :

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4
6
3
1

Answer :B
39.

In alpha-helix structure of protein, each turn has nearly ………………….. number of amino acids.

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3.6
5
4
2.7

Answer :A
40.

In alpha- helix structure intramolecular hydrogen bonding takes place between

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NH- group of one UNIT and gtCO group of ANOTHER DIFFERENT unit
NH- group of one unit and gtCO group of same part of same unit
No HYDROGEN bonding between-NH- group and gtCO group

Answer :B
41.

In almost all Indian metropolitan cities like Delhi, the major atmospheric pollutants is/are

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SUSPENDED PARTICULATE MATTER (SPM)
Oxides of nitrogen
Carbon DIOXIDE and carbon monoxide
Oxides of sulphur

Answer :A
42.

Inalpha -helix , structure, polypepetide chains are folded in a

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RIGHT HAND SIDE
LEFT hand side
both way
none of these.

ANSWER :A
43.

In allylic alcohol - OH group is attached to

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SP - hybridised CARBON ATOM
`sp^2` - hybridised carbon atom
`sp^3` - hybridised carbon atom
`sp^3`-d-hybridised carbon atom

Answer :C
44.

In all expected compounds each central atom only uses its s and p-orbitals in hybridization. The relationship between bond angle 'theta' and decimal fraction of s and p character present in the equivalent orbitals is given by : costheta=(S)/(S-1)=(P-1)/(P), S=decimal fraction of s-character in the equivalent hybrid orbitaland P=decimal fraction of p-character in the equivalent hybrid orbital. . Q.The incorrect statement is :

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The ratio in % p-character to % s-character is less than four, for the bond pair of CENTRAL atom of compound S
Central atom USES three hybrid orbitals to form COMPOUNDS R
Central atom uses four hybrid orbitals to form compounds S
There are three compounds present between point C to E, according to % s-character in bond pair of central atom.

Answer :D
45.

In all green leaves and vegetables which of the following vitamin is avialable ?

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VITAMIN
Vitamin D
Vitamin K
Vitamin `B_(12)`

ANSWER :C
46.

In all expected compounds each central atom only uses its s and p-orbitals in hybridization. The relationship between bond angle 'theta' and decimal fraction of s and p character present in the equivalent orbitals is given by : costheta=(S)/(S-1)=(P-1)/(P), S=decimal fraction of s-character in the equivalent hybrid orbitaland P=decimal fraction of p-character in the equivalent hybrid orbital. . Q. If the value n is 2 for compound T, then number of lone pair present at central atom of compound T will be :

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`0`
`1`
`2`
`3`

SOLUTION :
47.

In all expected compounds each central atom only uses its s and p-orbitals in hybridization. The relationship between bond angle 'theta' and decimal fraction of s and p character present in the equivalent orbitals is given by : costheta=(S)/(S-1)=(P-1)/(P), S=decimal fraction of s-character in the equivalent hybrid orbitaland P=decimal fraction of p-character in the equivalent hybrid orbital. . Q. The correct order of % p-character in bond pairs of central atoms in the following compounds is :

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`PgtTgtSgtQgtR`
`SgtRgtTgtPgtQ`
`PgtQgtSgtRgtT`
`PgtQgtSgtTgtR`

ANSWER :D
48.

In alkyl nitrites the oxygen of -O-N=O group is linked with carbon. An alkyl nitrite is :

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An ester
A NITRO compound
An amide
Anitrile

Answer :A
49.

In alkyl cyanide alkyl group attached with

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C of CN GROUP
N of CN group
Either C or N of CN group
Both C and N of CN group

Answer :A
50.

In alkanline medium ClO_(2) oxidises H_(2)O_(2) to O_(2) and itself gets reduced to Cl^(-). How many moles of H_(2)O_(2) are oxidised by 1 mol of ClO_(2)?

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`1.0`
`1.5`
`2.5`
`3.5`

Solution :`(({:(2ClO_(2)+2OH^(-)to2Cl^(-)+H_(2)O+5[O]),(H_(2)O_(2)+[O]toH_(2)O+O_(2)xx5):})/(5H_(2)O_(2)+2ClO_(2)+2OH^(-)to6H_(2)O+2Cl^(-)+5O_(2)))/`
THUS, ONE mole of `ClO_(2)` oxidises 5/2 or 2.5 moles of `H_(2)O_(2)`